Marbles problem to solve

Here's a nice "brain teaser" problem sent to my way... probably a student who wants to know the solution. I'll post it for you all and the solution in a few days.

204 marbles are divided into 3 groups according to colour. Ahmad found that there are twice as many blue marbles as white marbles and there are fewer red marbles than blue marbles. Ben found that the number of marbles in each group are divisible by 4 and 6. Cally found that the number of marbles in each group is less than 100.

How many red marbles are there?

(I do not have a clue where the problem originates, but I DO LIKE it!)

Solution:

Now, we need to find three numbers.

The one hint tells us that one number is double the other, and the third is less than the first (the doubled one). This won't yet get us started.

It is actually the latter hints that provide a starting point.

We learn the numbers are divisible by 4. They're also divisible by 6, which means they're divisible by 2 and 3. But we already knew they're divisible by 2 (since thye were divisible by 4), so the new information in this hint really is that the numbers are divisible by 3.

Divisible by 4 AND divisible by 3 means.... DIVISIBLE by 12!

Also all numbers are less than 100.

This really now restricts our "search space". We're looking at multiples of 12 that are less than 100.

Now, one was double the other. Since the 8th multiple of 12 (96) is the largest we can use, then two of the numbers COULD be the 4th and 8th multiples of 12 (48 and 96). They could also be the 3rd and 6th multiples of 12 (36 and 72).

The first hints will now "lock in" the solution... the sum has to be 204, which is 17 x 12. So, choosing 4 x 12, 8 x 12, and 5 x 12 - or 48, 96, and 60 - as our numbers will work. And, there are therefore 60 red marbles.

i loved it too and i have a solution1st B; nb of blue ones R: nd of red ones W: nd of white onesclearly b+r+w=204 divide by 12 the equation b1+r1+w1=17 where b1=b/12 and r1=r/12 and w1=w/12 and b1, r1 and w1 are natural since b, r and w divisble by 12now we know b1=2w1 since blure double that whiteso we got 3w1+r1=17 then w1=(17-r1)/3so we have 2 possible casses for r1 either 2 or 5 in order to get w1 natural number

1st case r1=2 then w1=5 and b1=10so r=24 w= 60 blue=100 which is rejected since all numbers are less than 100

Popular posts from this blog

I was just sent a link to this site; all it is, is a handy one-page printable conversion chart for various US measures, metric measures, and US vs. metric measures. Includes even a comparative Fahrenheit vs. Celsius thermometer.

Here's an enjoyable math video for students and parents/teachers alike. It's about folding and cutting... but it still IS very much about math -- because the lady models the basics of mathematical thinking in a wonderful way as she goes on cutting shapes... 😊

People sometimes ask me of my opinion or review of Saxon math. What I've written here applies in particular to Saxon Math's high school courses and middle grade levels. (The grades K-3 are by a different author and are quite different; more on that below.)

Saxon Math uses an "incremental approach" where math concepts are studied in little pieces over several lessons, and those lessons are strawed over a long period of time, intermixed with lessons about other topics.

In other words, if one lesson is on some particular topic (say, percentages or inequalities), it's almost guaranteed that the NEXT lesson is NOT on that topic. It jumps around from topic to topic constantly, and this is by design.

Saxon's method also includes a feature where after a lesson is taught, there are very few practice problems about the topic of the lesson. Most of the problems are mixed review problems, and they practice concepts from earlier lessons, not the concept …